johncarlosbaez

nancycomics, 5 hours ago to random

Nancy By Ernie Bushmiller May 22,1968

hanse_mina, 4 days ago to Ukraine

#Estonia's Prime Minister Kaja #Kallas was asked what was preventing the free world from doing more to help #Ukraine:

"Fear keeps us from supporting Ukraine. Countries have different fears, be it nuclear fear, fear of escalation, fear of migration. We must not fall into the trap of fear because that is what #Putin wants. He wants us to be afraid and not support Ukraine out of fear."

"Let's be decisive and not let our adversaries dictate our future,"

https://news.err.ee/1609346055/kallas-at-lennart-meri-conference-we-must-not-fall-into-the-trap-of-fears

#Russia

JimPropp, 5 days ago to random

My latest essay explains 0^0 = 1 by saying that the left-hand side is counting something that there's only 1 of: functions from the empty set to itself. To understand that, we need to look at what it might mean to add, multiply, and exponentiate sets. https://mathenchant.wordpress.com/2024/05/17/zero-to-the-zero-and-the-do-nothing-machine/

gregeganSF, 9 days ago to random

Wow, Thomas Hales and Koundinya Vajjha have proved Mahler’s First Conjecture! (That’s Kurt Mahler, not Gustave.)

https://arxiv.org/abs/2405.04331

Mahler’s First Conjecture says that the centrally symmetric convex shape with the worst possible packing ratio is made up of straight lines and arcs of hyperbolas, like the smoothed octagons shown in the animation below (demonstrating a 1-parameter family of slightly different packings all with the same density).

Whether these smoothed octagons are, as Karl Reinhardt conjectured, the actual worst case is still an open problem.

A bit more detail on Reinhardt’s conjecture in this article by @johncarlosbaez :

https://blogs.ams.org/visualinsight/2014/11/01/packing-smoothed-octagons/

and this thread by Koundinya Vajjha on Twitter:

https://twitter.com/KodyVajjha/status/1790211313618362848

A 1-parameter family of packings of smoothed octagons.

markmccaughrean, 12 days ago to Netherlands

Not bad; not bad at all 🙂

Especially for 52°N 🤷‍♂️

#AuroraSelfie
#Wassenaar
#Netherlands 🇳🇱
#Photography 📷
#SpaceWeather 🚀☁️
#AuroraBorealis

davidsuculum, 13 days ago to random

The association of Stack Overflow and chatGPT is exactly a proof of what it has already been said: the knowledge created by many people for free and to help others will be served to you for the cost of a premium account.

ErikJonker, 15 days ago to usa

This is a problem, for the first time in the history of the United States, billionaires had a lower effective tax rate than working-class Americans. The same problem also applies to countries in europe. Rich people can too easily avoid taxes. (giftarticle)
#NYT #taxrates #USA #billionaires #taxation
https://www.nytimes.com/interactive/2024/05/03/opinion/global-billionaires-tax.html?unlocked_article_code=1.qE0.1Zn-.suPDwq-AU_ox&smid=url-share

johncarlosbaez.wordpress.com, 18 days ago to random

https://commons.wikimedia.org/wiki/File:H3_633_FC_boundary.png

This picture by Roice Nelson shows a remarkable structure: the hexagonal tiling honeycomb.

What is it? Roughly speaking, a honeycomb is a way of filling 3d space with polyhedra. The most symmetrical honeycombs are the ‘regular’ ones. For any honeycomb, we define a flag to be a chosen vertex lying on a chosen edge lying on a chosen face lying on a chosen polyhedron. A honeycomb is regular if its geometrical symmetries act transitively on flags.

The most familiar regular honeycomb is the usual way of filling Euclidean space with cubes. This cubic honeycomb is denoted by the symbol {4,3,4}, because a square has 4 edges, 3 squares meet at each corner of a cube, and 4 cubes meet along each edge of this honeycomb. We can also define regular honeycombs in hyperbolic space. For example, the order-5 cubic honeycomb is a hyperbolic honeycomb denoted {4,3,5}, since 5 cubes meet along each edge:

Coxeter showed there are 15 regular hyperbolic honeycombs. The hexagonal tiling honeycomb is one of these. But it does not contain polyhedra of the usual sort! Instead, it contains flat Euclidean planes embedded in hyperbolic space, each plane containing the vertices of infinitely many regular hexagons. You can think of such a sheet of hexagons as a generalized polyhedron with infinitely many faces. You can see a bunch of such sheets in the picture:

https://commons.wikimedia.org/wiki/File:H3_633_FC_boundary.png

The symbol for the hexagonal tiling honeycomb is {6,3,3}, because a hexagon has 6 edges, 3 hexagons meet at each corner in a plane tiled by regular hexagons, and 3 such planes meet along each edge of this honeycomb. You can see that too if you look carefully.

A flat Euclidean plane in hyperbolic space is called a horosphere. Here’s a picture of a horosphere tiled with regular hexagons, yet again drawn by Roice:

Unlike the previous pictures, which are views from inside hyperbolic space, this uses the Poincaré ball model of hyperbolic space. As you can see here, a horosphere is a limiting case of a sphere in hyperbolic space, where one point of the sphere has become a ‘point at infinity’.

Be careful. A horosphere is intrinsically flat, so if you draw regular hexagons on it their internal angles are

2pi/3 = 120^circ

as usual in Euclidean geometry. But a horosphere is not ‘totally geodesic’: straight lines in the horosphere are not geodesics in hyperbolic space! Thus, a hexagon in hyperbolic space with the same vertices as one of the hexagons in the horosphere actually bulges out from the horosphere a bit — and its internal angles are less than 2pi/3: they are

arccosleft(-frac{1}{3}right) approx 109.47^circ

This angle may be familar if you’ve studied tetrahedra. That’s because each vertex lies at the center of a regular tetrahedron, with its four nearest neighbors forming the tetrahedron’s corners.

It’s really these hexagons in hyperbolic space that are faces of the hexagonal tiling honeycomb, not those tiling the horospheres, though perhaps you can barely see the difference. This can be quite confusing until you think about a simpler example, like the difference between a cube in Euclidean 3-space and a cube drawn on a sphere in Euclidean space.

Connection to special relativity

There’s an interesting connection between hyperbolic space, special relativity, and 2×2 matrices. You see, in special relativity, Minkowski spacetime is mathbb{R}^4 equipped with the nondegenerate bilinear form

(t,x,y,z) cdot (t',x',y',z') = t t' - x x' - y y' - z z

usually called the Minkowski metric. Hyperbolic space sits inside Minowski spacetime as the hyperboloid of points mathbf{x} = (t,x,y,z) with mathbf{x} cdot mathbf{x} = 1 and t > 0. But we can also think of Minkowski spacetime as the space mathfrak{h}_2(mathbb{C}) of 2×2 hermitian matrices, using the fact that every such matrix is of the form

A = left( begin{array}{cc} t + z & x - i y \ x + i y & t - z end{array} right)

and

det(A) = t^2 - x^2 - y^2 - z^2

In these terms, the future cone in Minkowski spacetime is the cone of positive definite hermitian matrices:

left{A in mathfrak{h}_2(mathbb{C}) , vert , det A > 0, , mathrm{tr}(A) > 0 right}

Sitting inside this we have the hyperboloid

mathcal{H} = left{A in mathfrak{h}_2(mathbb{C}) , vert , det A = 1, , mathrm{tr}(A) > 0 right}

which is none other than hyperbolic space!

Connection to the Eisenstein integers

Since the hexagonal tiling honeycomb lives inside hyperbolic space, which in turn lives inside Minkowski spacetime, we should be able to describe the hexagonal tiling honeycomb as sitting inside Minkowski spacetime. But how?

Back in 2022, James Dolan and I conjectured such a description, which takes advantage of the picture of Minkowski spacetime in terms of 2×2 matrices. And this April, working on Mathstodon, Greg Egan and I proved this conjecture!

I’ll just describe the basic idea here, and refer you elsewhere for details.

The Eisenstein integers mathbb{E} are the complex numbers of the form

a + b omega

where a and b are integers and omega = exp(2 pi i/3) is a cube root of 1. The Eisenstein integers are closed under addition, subtraction and multiplication, and they form a lattice in the complex numbers:

https://math.ucr.edu/home/baez/mathematical/eisenstein_integers.png

Similarly, the set mathfrak{h}_2(mathbb{E}) of 2×2 hermitian matrices with Eisenstein integer entries gives a lattice in Minkowski spacetime, since we can describe Minkowski spacetime as mathfrak{h}_2(mathbb{C}).

Here’s the conjecture:

Conjecture. The points in the lattice mathfrak{h}_2(mathbb{E}) that lie on the hyperboloid mathcal{H} are the centers of hexagons in a hexagonal tiling honeycomb.

Using known results, it’s relatively easy to show that there’s a hexagonal tiling honeycomb whose hexagon centers are all points in mathfrak{h}_2(mathbb{E}) cap mathcal{H}. The hard part is showing that every point in mathfrak{h}_2(mathbb{E}) cap mathcal{H} is a hexagon center. Points in mathfrak{h}_2(mathbb{E}) cap mathcal{H} are the same as 4-tuples of integers obeying an inequality (the mathrm{tr}(A) > 0 condition) and a quadratic equation (the det(A) = 1 condition). So, we’re trying to show that all 4-tuples obeying those constraints follow a very regular pattern.

Here are two proofs of the conjecture:

• John Baez, Line bundles on complex tori (part 5), The n-Category Café, April 30, 2024.

Greg Egan and I came up with the first proof. The basic idea was to assume there’s a point in mathfrak{h}_2(mathbb{E}) cap mathcal{H} that’s not a hexagon center, choose one as close as possible to the identity matrix, and then construct an even closer one, getting a contradiction. Shortly thereafter, someone on Mastodon by the name of Mist came up with a second proof, similar in strategy but different in detail. This increased my confidence in the result.

What’s next?

Something very similar should be true for another regular hyperbolic honeycomb, the square tiling honeycomb:

https://commons.wikimedia.org/wiki/File:H3_443_FC_boundary.png

Here instead of the Eisenstein integers we should use the Gaussian integers, mathbb{G}, consisting of all complex numbers

a + b i

where a and b are integers.

Conjecture. The points in the lattice mathfrak{h}_2(mathbb{G}) that lie on the hyperboloid mathcal{H} are the centers of squares in a square tiling honeycomb.

I’m also very interested in how these results connect to algebraic geometry! I explained this in some detail here:

• Line bundles on complex tori (part 4), The n-Category Café, April 26, 2024.

Briefly, the hexagon centers in the hexagonal tiling honeycomb correspond to principal polarizations of the abelian variety mathbb{C}^2/mathbb{E}^2. These are concepts that algebraic geometers know and love. Similarly, if the conjecture above is true, the square centers in the square tiling honeycomb will correspond to principal polarizations of the abelian variety mathbb{C}^2/mathbb{G}^2. But I’m especially interested in interpreting the other features of these honeycombs — not just the hexagon and square centers — using ideas from algebraic geometry.

https://johncarlosbaez.wordpress.com/2024/05/04/hexagonal-tiling-honeycomb/

nancycomics, 18 days ago to random

Mood…

randahl, 20 days ago to random

I have lost count of how many times, Russian media has claimed the sanctions were not working. But here is the hard truth:

The state owned energy company Gazprom had a net profit of USD 12,900,000,000 in 2022.

In 2023 the same company suffered a net loss of USD 6,900,000,000 — its first loss in 20 years.

Putin will lose this war.

Source: https://www.aljazeera.com/economy/2024/5/2/gazprom-falls-to-first-annual-loss-in-20-years-as-trade-with-europe-hit

SJHoodlet, 20 days ago to wildlife

Look what I found while moving rocks in my yard! 😍

"Oh, hello there!" I said after I squealed with joy.

When I was a child, I used to go hunting for newts, salamanders, snakes, turtles—you name it. It's been almost 20 years since I've seen one of these, and it was just as magical today as it was back then. 🥰

I placed it next to where I found it, where it crawled under a pile of loose soil and decaying leaf litter.

Until next time, my friend!

#Salamander #Wildlife #Nature #Spring

hanse_mina, 22 days ago to Russia

A wave of Ukrainian drone strikes on oil refineries deep inside #Russia has left the #Kremlin racing to defend its own territory while still waging war on its neighbor. But the attacks have also achieved the unthinkable — leaving the world’s largest petrostate running low on petrol.

Diesel prices for Russian consumers have skyrocketed, rising almost 10 percent in the past week alone.

https://www.politico.eu/article/vladimir-putin-russia-diesel-prices-skyrocket-ukraine-war-drone-strikes-oil-refineries/

#Ukraine

Joemoeller, 22 days ago to random

This summer I'm starting a postdoc at CalTech! I'm working with the roboticist/control theorist Aaron Ames on using category theory to study stability of dynamical systems.

I got verbal confirmation a month ago, but I've been holding back on saying anything until I got bureaucratic confirmation.

MartinEscardo, 25 days ago to random

What is a topological space?

It is a mathematical device to define what a continuous function is, in a general setting.

1. A topological space is a set X together with a collection of subsets of X, called open, such that finite intersections of open sets are open, and arbitrary unions of open sets are open.

2. A function of topological spaces is continuous if inverse images of open sets are open.

What is the intuition behind (1) and (2)?

I claim that it is better to ask, instead, how mathematicians came up with (1) and (2).

1/

marcioaleks, 26 days ago to random

Three-Body Problem. Newly discovered stable periodic orbits.

By Xiaoming LI and Shijun LIAO
Source: https://numericaltank.sjtu.edu.cn/three-body/three-body.htm

video/mp4

GottaLaff, 28 days ago to random

Via Acyn (who has the vid):

#Biden: "Remember when #Trump was trying to deal with COVID, he suggested injecting a little bleach in your veins… he missed, it all went to his hair (Loud, extended laughter and applause)

(smiling) I shouldn't have said that... You guys are a bad influence on me."

paulbalduf, 27 days ago to physics

Here is a curious finding from our statistical analysis https://arxiv.org/abs/2403.16217 :
A #Feynmangraph is a graphical short hand notation for a complicated integral that computes the probability for scattering processes in #quantum field theory.
An electrical circuit can also be described as a graph. What happens if we interpret the Feynman graph as an #electrical network, where each edge is a 1 Ohm resistor? We can then compute the resistance between any pair of vertices and collect all these values in a "resistance matrix", as shown below. The average of all these resistances is called "Kirchhoff index". Now it turns out that this average resistance is correlated fairly strongly with the Feynman integral of that graph: A graph with large contribution to quantum scattering amplitudes on average also has a large electrical resistance. Isn't that a nice connection between two seemingly distinct branches of theoretical #physics ?

Correlation between average resistance and Feynman period

lana, 29 days ago to random

Exceedingly cool results.
Somehow, wherever this small bat thrives, its presence also creates a niche for a bigger bat, which evolves from... the small bat. You get a string of islands with the simultaneous presence of the small bat and the big bat that re-evolves from it every time....

The phenomenon is so weird that people thought it was just 2 different species that colonized the islands in parallel. It's just one species with a recurring sub-branch.

https://phys.org/news/2024-04-species-sizes-rare-evolution-action.html

johncarlosbaez, 1 month ago (edited 1 month ago) to random

The 'hexagonal tiling honeycomb' is a beautiful structure in 3-dimensional hyperbolic space. I'm trying to figure out something about it.

It contains infinitely many sheets of hexagons, tiling planes in the usual way hexagons do. These are flat Euclidean planes in 3d hyperbolic space, called 'horospheres'. I want to know the coordinates of the vertices of these hexagons. I have some clues.

The hexagonal tiling honeycomb has Schläfli symbol {6,3,3} . The Schläfli symbol is defined in a recursive way. The symbol for the hexagon is {6}. The symbol for the hexagonal tiling of the plane is {6,3} because 3 hexagons meet at each vertex. Finally, the hexagonal tiling honeycomb has symbol {6,3,3} because 3 hexagonal tilings meet at each edge!

The symmetry group of the hexagonal tiling honeycomb is the Coxeter group {6,3,3}. This is a discrete subgroup of the Lorentz group O(3,1), which acts on 3d hyperbolic space because that space is the set of points (t,x,y,z) in Minkowski spacetime with

t² − x² − y² − z² = 1 and t > 0

The Coxeter group {6,3,3} is generated by reflections, but its 'even part', generated by pairs of reflections, is a discrete subgroup of PSL(2,ℂ), because this is the identity component of the Lorentz group. In fact, this Coxeter group is almost PSL(2,𝔼), where 𝔼 is the ring of 'Eisenstein integers'. These are complex numbers of the form

a + bω

where a,b are integers and ω is a nontrivial cube root of 1. So there should be a nice description of the hexagonal tiling honeycomb using Eisenstein integers! And this is what I'm trying to find... quickly, before May 1st because I'm have a column due then. 😧

I should ask @roice3, who drew this....

(1/n)

gsuberland, 1 month ago (edited 1 month ago) to random

discovering that fewer cybertrucks have been sold than Sinclair C5s has amused me greatly. and honestly, if you want to drive a weird vehicle, the C5 is far more fun.

https://en.wikipedia.org/wiki/Sinclair_C5

GottaLaff, 1 month ago to random

Via Jesse Rodriguez:

#Trump has to be in the courtroom every day for the duration of his criminal trial.

"If you do not show up there will be an arrest," Judge Merchan said. #TrumpTrial

ProfKinyon, 1 month ago to random

Theorem: If N is a positive integer and is not a perfect square, then (\sqrt{N}) is irrational.

Proof: Suppose (\sqrt{N} = a/b) for positive integers (a,b) with no common factor greater than 1. Then (b/a = \sqrt{N}/N), and so (a/b = (bN)/a). Since the first fraction is in lowest terms, the numerator and denominator of the second fraction must be a common integer multiple, say (c), of the numerator and denominator of the first. Hence (a = cb), and therefore, (\sqrt{N} = c), that is, (N) is a perfect square. QED

I learned this proof from a one paragraph insert in the American Mathematical Monthly (vol. 115, June-July 2008, p. 524) written by Geoffrey C. Berresford. I just love it.

mariescopy, 1 month ago to evolution

Very interesting story about the #nitroplast, a potential new organelle resulting from #symbiosis between nitrogen-fixing bacteria and the #microalgae Braarudosphaera bigelowii based on subcellular imaging and proteomic data.
Possibly a new model to study the evolutionary transition from #endosymbiont to organelle.
https://www.science.org/doi/10.1126/science.adk1075

With a Perspective comment by Ramon Massana
https://www.science.org/doi/10.1126/science.ado8571
#evolution #protists

coreyspowell, 1 month ago (edited 1 month ago) to space

Today's dose of cosmic beauty: Comet Pons-Brooks, photographed from southern Spain by Fritz Helmut Hemmerich.

The eerie green glow is from diatomic carbon. The ripples show the flow of the solar wind.

https://www.facebook.com/fritzhelmut.hemmerich/ #space #science #astronomy #nature

TodayInTwitter, 1 month ago to random

X rolled out a change that automatically converts all links that have the string “twitter” anywhere in them to “X.”

For example, if someone tweets the link setwitter.com, the X client application will change that link to sex.com.

This is not a joke.

https://mashable.com/article/twitter-dot-com-posts-change-to-x-dot-com-ios